Question: $ \left(\dfrac{9}{4}\right)^{-\frac{3}{2}}$
Answer: $= \left(\dfrac{4}{9}\right)^{\frac{3}{2}}$ $= \left(\left(\dfrac{4}{9}\right)^{\frac{1}{2}}\right)^{3}$ To simplify $\left(\dfrac{4}{9}\right)^{\frac{1}{2}}$ , figure out what goes in the blank: $\left(? \right)^{2}=\dfrac{4}{9}$ To simplify $\left(\dfrac{4}{9}\right)^{\frac{1}{2}}$ , figure out what goes in the blank: $\left({\dfrac{2}{3}}\right)^{2}=\dfrac{4}{9}$ so $ \left(\dfrac{4}{9}\right)^{\frac{1}{2}}=\dfrac{2}{3}$ So $\left(\dfrac{4}{9}\right)^{\frac{3}{2}}=\left(\left(\dfrac{4}{9}\right)^{\frac{1}{2}}\right)^{3}=\left(\dfrac{2}{3}\right)^{3}$ $= \left(\dfrac{2}{3}\right)^{3}$ $= \left(\dfrac{2}{3}\right)\cdot\left(\dfrac{2}{3}\right)\cdot \left(\dfrac{2}{3}\right)$ $= \dfrac{4}{9}\cdot\left(\dfrac{2}{3}\right)$ $= \dfrac{8}{27}$